3.13 \(\int (a+a \sec (e+f x))^3 (c-c \sec (e+f x))^3 \, dx\)

Optimal. Leaf size=68 \[ -\frac{a^3 c^3 \tan ^5(e+f x)}{5 f}+\frac{a^3 c^3 \tan ^3(e+f x)}{3 f}-\frac{a^3 c^3 \tan (e+f x)}{f}+a^3 c^3 x \]

[Out]

a^3*c^3*x - (a^3*c^3*Tan[e + f*x])/f + (a^3*c^3*Tan[e + f*x]^3)/(3*f) - (a^3*c^3*Tan[e + f*x]^5)/(5*f)

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Rubi [A]  time = 0.07435, antiderivative size = 68, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.115, Rules used = {3904, 3473, 8} \[ -\frac{a^3 c^3 \tan ^5(e+f x)}{5 f}+\frac{a^3 c^3 \tan ^3(e+f x)}{3 f}-\frac{a^3 c^3 \tan (e+f x)}{f}+a^3 c^3 x \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sec[e + f*x])^3*(c - c*Sec[e + f*x])^3,x]

[Out]

a^3*c^3*x - (a^3*c^3*Tan[e + f*x])/f + (a^3*c^3*Tan[e + f*x]^3)/(3*f) - (a^3*c^3*Tan[e + f*x]^5)/(5*f)

Rule 3904

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.), x_Symbol] :> Di
st[(-(a*c))^m, Int[Cot[e + f*x]^(2*m)*(c + d*Csc[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x]
&& EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[m] && RationalQ[n] &&  !(IntegerQ[n] && GtQ[m - n, 0])

Rule 3473

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(b*Tan[c + d*x])^(n - 1))/(d*(n - 1)), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

\begin{align*} \int (a+a \sec (e+f x))^3 (c-c \sec (e+f x))^3 \, dx &=-\left (\left (a^3 c^3\right ) \int \tan ^6(e+f x) \, dx\right )\\ &=-\frac{a^3 c^3 \tan ^5(e+f x)}{5 f}+\left (a^3 c^3\right ) \int \tan ^4(e+f x) \, dx\\ &=\frac{a^3 c^3 \tan ^3(e+f x)}{3 f}-\frac{a^3 c^3 \tan ^5(e+f x)}{5 f}-\left (a^3 c^3\right ) \int \tan ^2(e+f x) \, dx\\ &=-\frac{a^3 c^3 \tan (e+f x)}{f}+\frac{a^3 c^3 \tan ^3(e+f x)}{3 f}-\frac{a^3 c^3 \tan ^5(e+f x)}{5 f}+\left (a^3 c^3\right ) \int 1 \, dx\\ &=a^3 c^3 x-\frac{a^3 c^3 \tan (e+f x)}{f}+\frac{a^3 c^3 \tan ^3(e+f x)}{3 f}-\frac{a^3 c^3 \tan ^5(e+f x)}{5 f}\\ \end{align*}

Mathematica [A]  time = 0.0401183, size = 61, normalized size = 0.9 \[ -a^3 c^3 \left (\frac{\tan ^5(e+f x)}{5 f}-\frac{\tan ^3(e+f x)}{3 f}-\frac{\tan ^{-1}(\tan (e+f x))}{f}+\frac{\tan (e+f x)}{f}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sec[e + f*x])^3*(c - c*Sec[e + f*x])^3,x]

[Out]

-(a^3*c^3*(-(ArcTan[Tan[e + f*x]]/f) + Tan[e + f*x]/f - Tan[e + f*x]^3/(3*f) + Tan[e + f*x]^5/(5*f)))

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Maple [A]  time = 0.023, size = 93, normalized size = 1.4 \begin{align*}{\frac{1}{f} \left ( -3\,{c}^{3}{a}^{3}\tan \left ( fx+e \right ) +{c}^{3}{a}^{3} \left ( fx+e \right ) -3\,{c}^{3}{a}^{3} \left ( -2/3-1/3\, \left ( \sec \left ( fx+e \right ) \right ) ^{2} \right ) \tan \left ( fx+e \right ) +{c}^{3}{a}^{3} \left ( -{\frac{8}{15}}-{\frac{ \left ( \sec \left ( fx+e \right ) \right ) ^{4}}{5}}-{\frac{4\, \left ( \sec \left ( fx+e \right ) \right ) ^{2}}{15}} \right ) \tan \left ( fx+e \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sec(f*x+e))^3*(c-c*sec(f*x+e))^3,x)

[Out]

1/f*(-3*c^3*a^3*tan(f*x+e)+c^3*a^3*(f*x+e)-3*c^3*a^3*(-2/3-1/3*sec(f*x+e)^2)*tan(f*x+e)+c^3*a^3*(-8/15-1/5*sec
(f*x+e)^4-4/15*sec(f*x+e)^2)*tan(f*x+e))

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Maxima [A]  time = 1.03518, size = 127, normalized size = 1.87 \begin{align*} -\frac{{\left (3 \, \tan \left (f x + e\right )^{5} + 10 \, \tan \left (f x + e\right )^{3} + 15 \, \tan \left (f x + e\right )\right )} a^{3} c^{3} - 15 \,{\left (\tan \left (f x + e\right )^{3} + 3 \, \tan \left (f x + e\right )\right )} a^{3} c^{3} - 15 \,{\left (f x + e\right )} a^{3} c^{3} + 45 \, a^{3} c^{3} \tan \left (f x + e\right )}{15 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(f*x+e))^3*(c-c*sec(f*x+e))^3,x, algorithm="maxima")

[Out]

-1/15*((3*tan(f*x + e)^5 + 10*tan(f*x + e)^3 + 15*tan(f*x + e))*a^3*c^3 - 15*(tan(f*x + e)^3 + 3*tan(f*x + e))
*a^3*c^3 - 15*(f*x + e)*a^3*c^3 + 45*a^3*c^3*tan(f*x + e))/f

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Fricas [A]  time = 1.06599, size = 189, normalized size = 2.78 \begin{align*} \frac{15 \, a^{3} c^{3} f x \cos \left (f x + e\right )^{5} -{\left (23 \, a^{3} c^{3} \cos \left (f x + e\right )^{4} - 11 \, a^{3} c^{3} \cos \left (f x + e\right )^{2} + 3 \, a^{3} c^{3}\right )} \sin \left (f x + e\right )}{15 \, f \cos \left (f x + e\right )^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(f*x+e))^3*(c-c*sec(f*x+e))^3,x, algorithm="fricas")

[Out]

1/15*(15*a^3*c^3*f*x*cos(f*x + e)^5 - (23*a^3*c^3*cos(f*x + e)^4 - 11*a^3*c^3*cos(f*x + e)^2 + 3*a^3*c^3)*sin(
f*x + e))/(f*cos(f*x + e)^5)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - a^{3} c^{3} \left (\int \left (-1\right )\, dx + \int 3 \sec ^{2}{\left (e + f x \right )}\, dx + \int - 3 \sec ^{4}{\left (e + f x \right )}\, dx + \int \sec ^{6}{\left (e + f x \right )}\, dx\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(f*x+e))**3*(c-c*sec(f*x+e))**3,x)

[Out]

-a**3*c**3*(Integral(-1, x) + Integral(3*sec(e + f*x)**2, x) + Integral(-3*sec(e + f*x)**4, x) + Integral(sec(
e + f*x)**6, x))

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Giac [A]  time = 1.47474, size = 93, normalized size = 1.37 \begin{align*} -\frac{3 \, a^{3} c^{3} \tan \left (f x + e\right )^{5} - 5 \, a^{3} c^{3} \tan \left (f x + e\right )^{3} - 15 \,{\left (f x + e\right )} a^{3} c^{3} + 15 \, a^{3} c^{3} \tan \left (f x + e\right )}{15 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(f*x+e))^3*(c-c*sec(f*x+e))^3,x, algorithm="giac")

[Out]

-1/15*(3*a^3*c^3*tan(f*x + e)^5 - 5*a^3*c^3*tan(f*x + e)^3 - 15*(f*x + e)*a^3*c^3 + 15*a^3*c^3*tan(f*x + e))/f